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The formula for Heisenberg's uncertainty...

The formula for Heisenberg's uncertainty principle is

A

`lambda = ( h)/( mv)`

B

`Deltax xx Deltap ge ( h)/( 4pi)`

C

`Deltax xx Deltap ge ( h)/( 2pi)`

D

`mvr = n (h)/(2pi)`

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The correct Answer is:
**Step-by-Step Solution:** 1. **Understanding Heisenberg's Uncertainty Principle**: The Heisenberg's uncertainty principle states that it is impossible to simultaneously know both the position and momentum of a particle with absolute precision. This principle is fundamental in quantum mechanics. 2. **Identifying the Variables**: In the context of the uncertainty principle: - \( \Delta x \) represents the uncertainty in position. - \( \Delta p \) represents the uncertainty in momentum. 3. **Writing the Mathematical Expression**: The mathematical expression for Heisenberg's uncertainty principle is: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] where \( h \) is Planck's constant. 4. **Analyzing the Options**: The given options are: - a) \( \lambda = \frac{h}{mv} \) (De Broglie wavelength) - b) \( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \) (Correct) - c) \( \Delta x \cdot \Delta p \geq \frac{h}{2\pi} \) (Incorrect) - d) \( mv = \frac{nh}{2\pi} \) (Quantization of angular momentum) 5. **Selecting the Correct Answer**: Based on the analysis, option b) \( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \) is the correct formula for Heisenberg's uncertainty principle. **Final Answer**: The formula for Heisenberg's uncertainty principle is \( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \). ---
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Mathematically, Heisenberg's uncertainty principle can best be explained by

It is impossible to determine simultaneously the position of velocity of small microscopic particle such as electron , proton or neutron with accuracy .This is called Heisenberg's uncertainty principle. Mathematically, it is represented as Delta x. Delta p ge (h)/(4pi) , Delta x is uncertainty in position Delta p is uncertainty in momentum.

Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse also true. This is summed up in what we now call the Heisenberg uncertainty principle. The equation si deltax.delta (mv)ge(h)/(4pi) The uncertainty in the position or in the momentum of a marcroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electon is small enough for the uncertainty to be relatively large and significant. If the uncertainties in position and momentum are equal, the uncertainty in the velocity is :

AAKASH INSTITUTE ENGLISH-STRUCTURE OF ATOM -ASSIGNMENT ( SECTION -A) Objective Type Questions (One option is correct)
  1. de-Broglie wavelength applies only to

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  2. The wavelength associated with an electron moving with velocity 10^(10...

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  3. The formula for Heisenberg's uncertainty principle is

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  4. Who first time ruled out the existence of definite paths of electrons ...

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  5. Probability density is given by

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  6. The possible values of magnetic quantum number for p-orbital are

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  7. The notation of orbital with n=5 and l=3 is

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  8. The possible values of m(l) in a given subshell are given by the formu...

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  9. The shape of p orbital is

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  10. In multi-electron atom 4s-orbital is lower in energy than

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  11. The main energy shell in which the electron is present is given by

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  12. Shape of an orbital is given by

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  13. Orientation of orbitals is given by

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  14. Which one of the following orbitals is spherical in shape ?

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  15. The number of valence electrons in Aluminium is

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  16. For n=4 , which one of the following values of l is not possible ?

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  17. The number of electrons present in 3d of Cu^(o+) is

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  18. The maximum number of electrons that can be accomoddated in d(x^(2)-y^...

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  19. The number of unpaired electrons in magnesium atom is

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  20. The correct sequence of energy of orbitals of multielectron species i...

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